
Introduction:
In this project we will design a parking umbrella that will fit two cars at the same time; it
will be designed such as a truss. This umbrella is used to shade cars in the summer from
sunlight or cover cars from rain and snow.
We will design the truss to make it easily assembled on the site and can be easily stored at
the storage place as a pre made kit, also minimizing the cost of the truss and its
manufacturing. As for designing for Jordan, the truss should be carefully designed to take a
high load of snow accumulation.
-
 Constrains and Analysis:
Can cover two regular size cars.
Can handle accumulation of 30 cm, such that the snow has density of 12% of water.
Use a design factor of two ( nd = 2 ).
The weight of the snow is 40% at the top of the truss, 15% at each corner.
The maximum deflection between the truss and the ground to be 10 cm.
The clearance between the umbrella and the ground is 210 cm.
A
Sixty centimeters of the column length is buried in the ground.
-
 Force Analysis:
When we take the section A-A.
7.0711
5
0.2 F
0.15 F
0.373 F
A
0.15 F
3.66
15⁰
0.373 F
0.15 F
0.25 F
0.25 F
0.373 F
-
determine all the reactions and the internal forcesof the beam
0.25 F
If we take the top view:
side view:
0.5982 F
0.264 F
0.2794 F
0.264 F
Cable
0.25 F
0.264 F
0.386F
2.3
Loads on the column
0.586 F
2.1
0.25F
0.264F
0.6
We will have failure due to two buckling segment and due to bending
Moment on the C-channel,
0.5 F
1.804176𝐹
F
0.5 F
A
7.0711

Selection on the beams:
3
5
The beams are pointed as in the figure in numbers in red
1
A
assume all are pinned-pinned
-
Beam (1):
2
We selected from table A-17
Thick 40 with outer diameter 41.5 mm
As shown from the iteration table shown at the end of this report.
𝐴=
𝜋
(41.52 − (41.5 − (2 × 3.375)2 ) = 404.3𝑚𝑚2
4
𝜎𝑛 =
𝑃𝑐𝑟 =
0.279 × 8829
= 6.09 𝑀𝑃𝑎
404.3
𝐶𝜋 2 𝐸𝐼 2 × 𝜋 2 × 210 × 109 × 7.402 × 10−8
=
= 63136.66 𝑁
𝐿2
3.662
The factor of safety due to the compression in the beam:
𝑛=
The factor of safety due to buckling:
𝑆𝑦 310
=
= 50.87
𝜎𝑛 6.09
𝑛=
-
𝑃𝑐𝑟 63136.66
=
= 2.491
𝑃
2436.3
Beam (2):
The second beam will be selected as the beam number 1:
The factor of safety due to tension is:
𝑆𝑦 310
=
= 47.74
𝜎𝑛 6.49
𝑛=
-
Beam (3):
The third beam has a compression load so it might fail due to compression or buckling
We can select ( Extra Strong with outer diameter 32.875mm ) but its weight will be higher to
choose 41.5mm Thick 40.
Due to compression:
𝑛=
𝑆𝑦 310
=
= 36.77
𝜎𝑛 8.43
𝑛=
𝑃𝑐𝑟 11450
=
= 3.36
𝑃
3408
Due to buckling:
-
Cable:
In the cable there is only failure due to tension so we selected 16.875 mm Thick 40,
We made this selection due to practical considerations because the bolt will take a lot of the
pipe area and might lead to failure due to the holes of the pipe so we chosen this size.
𝑛=
-
𝑆𝑦 310
=
= 6.39
𝜎𝑛
49
Column:
We will choose a C-channel with dimensions (152 X 76)
The column as we mentioned before has two segment of failure due to buckling and one
due to normal stress from compression and bending moment. Also it can be put in two
orientations in each one the moment of inertia and the value of Z will vary changing the stress
and the Pcr value so we need to know the best orientation and size for the column.
So in the table of iterations we will see that the only expected mode of failure is due to
compression and bending (Normal Stress)
The maximum normal stress:
𝜎𝑁 =
𝑃 𝑀 0.5 × 8829 1.804176 × 8829
+ =
+
= 144.417 𝑀𝑃𝑎
𝐴 𝑍
22.7
111.8
𝑛=
-
𝑆𝑦
310
=
= 2.146
𝜎𝑛 144.417
Weld mints in the joints:
The weld is shown in brown and in red the projected area of
The weld , so we can approximate it as:
0.25 F
0.373 F
𝑉
𝜏𝑎𝑙𝑙 = 𝐴 =
0.45×8829
1.414𝜋ℎ𝑟
3973.05
= 0.07301ℎ =
393×106
2×1.44
0.45 F
ℎ = 0.3988𝑚𝑚
Which can be reached easily in the welding.
0.373 F
The second weld on the same joint:
We can approximate it as:
𝜏𝑎𝑙𝑙 =
𝑉 0.373 × 8829
3293.217
393 × 106
=
=
=
𝐴
1.414𝜋ℎ𝑟
00.07301ℎ
2 × 1.44
ℎ = 0.3055 𝑚𝑚
0.125
F
The weld mint in the column:
0.132
F
First we will calculate the primary and secondary shear
d
𝜏́ 1 =
𝑉 0.125 × 8829 31532
=
=
𝐴
0.7ℎ𝑑
ℎ
𝑉 0.132 × 8829 33298
𝜏́ 2 = =
=
𝐴
0.7ℎ𝑑
ℎ
0.125 F
2154331.9
ℎ
25 mm
𝐽𝑢 =
𝑑3
12
63.46 N.m
𝑀𝑟
1.58645
2154331.9
𝜏̈ =
=
=
𝐽
7.364 × 10−6 ℎ × 2
ℎ×2
𝜏𝑚𝑎𝑥
0.132 F
2154331.9
ℎ
50 mm
31532 2
33298 + 2154331.9 2
250728.636 393 × 106
√
= (
) +(
) =
=
ℎ×2
ℎ×2
ℎ×2
2 × 1.44
ℎ = 0.9187 𝑚𝑚
-
Weld mint in the in the metal sheet in the corner joint:
𝜏́ 1 =
𝑉 0.25 × 8829 71210.7
=
=
𝐴
0.7ℎ𝑑
ℎ
𝜏́ 2 =
𝑉 0.54347 × 8829 154803.73
=
=
𝐴
0.7ℎ𝑑
ℎ
𝜏̈ =
𝑀𝑟
2.81
549345.2
=
=
6
𝐽
5.1152 × 10 ℎ
ℎ
0.25F
0.54347 F
71210.7 549345.2 2
154803.73 2 639573.1316 393 × 106
𝜏 = √(
+
) +(
) =
=
ℎ
ℎ
ℎ
ℎ
2 × 1.44
ℎ = 0.4687 𝑚𝑚
126.9 N.m
44.28mm
-
Welding in the head joint:
It can be approximated as a circle
𝜏𝑎𝑙𝑙 =
0.4 F
𝑉 0.5873 × 8829
5185.69
393 × 106
=
=
=
𝐴
1.414𝜋ℎ𝑟
00.07301ℎ
2 × 1.44
ℎ = 0.5205 𝑚
0.5873 F
0.5459 F
-
weld mint in the column which the cable is set:
𝑉
0.125×8829
0.7ℎ𝑑
𝑉
0.2717×8829
0.7ℎ𝑑
𝑀𝑟
𝐽
1.5865
7.364×10−6 ℎ
𝜏́ 1 = 𝐴 =
𝜏́ 2 = 𝐴 =
𝜏̈ =
=
=
31532
ℎ
=
68538.26
ℎ
=
215440
ℎ
0.2991 F
F
2154331.9
ℎ
0.125 F
25 mm
63.46N.m
0.2717 F
50 mm
2
2
6
68538.26 215440
31532
285723.81 393 × 10
𝜏 = √(
+
) +(
) =
=
ℎ
ℎ
ℎ
ℎ
2 × 1.44
ℎ = 0.21 𝑚𝑚
-
The modes of failure from the bolt:
Bearing
𝐹=
𝑡𝑑𝑆𝑝
𝑛𝑑
Non-threaded:
𝐹=
0.577𝜋𝑑 2 𝑆𝑝
4𝑛𝑑
Threaded:
𝐹=
0.577𝐴𝑟 𝑆𝑝
𝑛𝑑
Tearing:
𝐹=
𝑎𝑡0.577𝑆𝑚𝑒𝑚
𝑛𝑑
Take any bolt from standard sizes and substitute in this equations and the lower force must
be equal the needed force. For example in the beam which has 0.279F load select 8mm size
3.375 × 2 × 8 × 310
𝐹=
= 8370𝑁
2
0.577 × 𝜋 × 82 × 310
𝐹=
= 8991𝑁
4×2
0.577 × 32.8 × 310
𝐹=
= 2933𝑁
2
But the force needed F = 0.279× 8829 = 2463.3N
So this bolt will use
2154331.9
ℎ
Defliction in the beam:
P3
P2
from Gastaceliano's theorem:
P1
2.1
2.1
4.4 (𝑃
𝑑𝑈
2(𝑃2 + 𝑃3 )𝑥 2
2 + 𝑃3 )𝑥 + (𝑃2 (𝑥 − 2.1))
=∫
𝑑𝑥 + ∫
𝑑𝑥
𝑑𝑃3
2𝐸𝐼
𝐸𝐼
0
2.1
𝛿 = 5.812 𝑐𝑚
2.3